Module 4

Extending the Ideas of LinearProgramming

4.1 Introduction

4.2 Deriving Additional Information from lP4.2.1 Dual Values4.2.2 Right-Hand Side Ranges4.2.3 Coefficient Ranges

4.3 Decision Problems with lP Associations

4.4 Integer Programming

4.5 Non-linear Programming

4.6 Programming under Uncertainty

4.7 Concluding Remarks

Review Questions

Case Study 4.1

Case Study 4.2

Case Study 4.3

4/1

4/24/24/44/6

4/9

4/13

4/13

4/15

4/16

4/17

4/19

4/19

4/20

This is an advanced module and may be omitted on the first time through thecourse. At the start of the module the reader should already be familiar withthe fundamentals of LP.By the end he or she should know how to extend thesebasic ideas, in terms both of generating further information and of applyingthem to problems that do not appear at first sight to be LP problems. Someunderlying assumptions on LP can be relaxed if methods such as integer andnon-linear programming are used.

Prerequisite Module

Module 3: Linear Programming

LP solves decision problems which require the optimisation of an objective suchas profit, but where the actions that can be taken are restricted by constraints,such as the amount of finance available. If the objective and the constraintscan be expressed mathematically in terms of decision variables, and if these

mathematical expressions'are'linear (they contain no logarithmic squared, cubed,etc., terms) then LP has the potential to solve this problem.This basic idea, as described in the previous module, is just the beginning.

The technique can produce much more information than the bare solution to aproblem; it can deal with a much wider set of problems than may be apparentat first sight; it can be adapted to solve problems for which the assumptions ofLP, such as linearity, are not met; it can solve problems in which some or allof the variables can take only whole number values; it can also be applied toproblems where there is uncertainty, in the sense that some decision variablesmay have a statistical distribution, for example when the variable describesweather conditions such as rainfall. This module describes all these differenttypes of extension to the concept of LP.

When a computer algorithm such as Simplex solves an LP problem, it provides,besides the optimum values of the decision variables and the objective func-tion, other information which can be used to perform sensitivity analysis. Theinformation can be used to test how the optimal solution would be affected ifthe basic assumptions or input data of the problem were changed. Included inthis information aredual values, sometimes called shadow prices. They refer toconstraints, not variables. Each constraint has a dual value which is a measure ofthe increase in the objective function if one extra unit of the resource associatedwith the constraint were available, everything else being unchanged.The electronics example used in the previous module was:

Maximise profitSubject to

4x +3y5x + 2y ~ 80 (available machine-hours)3x + 4y ~ 90 (available man-hours)

x,y ;::: aThe solution was: x = la, y = 15; objective function = 85The dual values, which would be given as part of the computer output, are:

Constraint Dual value5x + 2y ~ 80 0.5 (machine-hours)3x + 4y ~ 90 0.5 (man-hours)

The interpretation of the dual values is that if the available machine-hourswere 81 instead of 80, the objective function value could be increased by 0.5 to85.5. So, were the decision maker offered an extra machine-hour at the sameprice as the previous 80 (the price already included in the objective function) heor she should accept because using it can increase profit by 0.5. If the decisionmaker has to pay an extra amount for the extra unit, 0.5 is the maximum thatshould be paid.

An extra man-hour (91 available instead of 90) would also increase the objec-tive function value by 0.5 to 85.5. It is only by chance that the dual values areequal. Usually the dual values will differ from constraint to constraint. Dualvalues are measures of the true worth of resources to the decision maker andthere is no reason why they should be the same as the price paid for the resourceor its market value. The dual values are produced as part of the output of analgebraic solution of an LP problem and require no extra calculations once theLP has been solved.Dual values are sometimes equal to zero. In the example, both constraints

were tight, meaning that the total availability of each resource was used up.(Check this by substituting the optimal x, y values in the constraints.) In somecircumstances the total availability of a resource might not be fully used up.If not all the resource is used up the constraint is said to be slack and theassociated dual value (of that constraint only) must be zero. Since units of thatresource are available but not being employed, extra units of it have no valueto the decision maker: when the slack is non-zero then the dual value is zero.

A dual value can be calculated graphically by reworking the solution with theright-hand side of the constraint increased by 1 (see Figure 4.1). The amount bywhich the objective function has increased in the new solution is the dual value.In the example, the dual value of machinery can be calculated by solving theproblem as before, but with the machinery constraint being 5x + 2y ::;81. In thiscase, the optimal point is the point of intersection of:

5x + 2y = 813x + 4y = 90

The optimal point is no longer A, but is N. The x and y values can be calculatedby simultaneous equations as before, giving:

2x = 10-

7

11Y = 14-14

2 114(10-) + 3(14-) = 85.5

7 14

Dual value = 85.5 - 85= 0.5

At first sight it may seem paradoxical that an extra machine-hour could beused at all. In the original solution all 80 machine-hours and all 90 man-hourswere being used. Since both products X and Y require a combination of bothresources, how could the extra machine-hour be used since there are no man-hours with which to combine it? The answer is that the production mix changes.Fewer units of Yare made (y decreases from 15 to 14H). This frees up someunits of both resources which can be combined with the extra machine-hour toproduce more of product X (x increases from 10 to 10~). The new productionmix is more profitable by 0.5, as measured by the dual value.The dual value works in both directions. A dual value of 0.5 also means thatone unit less of the resource would lower the objective function by 0.5.

Dual values are marginal values and are not guaranteed to hold over largeranges. If the decision maker were offered ten extra machine-hours at an addi-tional price of 0.3, he should accept at least one since 0.3 is less than the dualvalue of 0.5. But should he accept all ten? Most computer packages give therange for which the dual value holds. This is known as the right-hand siderange.Again, a graph helps to show how a range is calculated. In Figure 4.2, as

the number of machine-hours available is increased, the constraint is movedoutwards. For the first few increased units the optimal point, the intersection ofthe two constraints, moves. The objective function increases just as it did as theoptimal point moved from A to N when the dual value was calculated. Justas for the dual value calculation each unit increase in the resource results inthe same 0.5 increase in the objective function. In other words the dual value isconstant at 0.5.

However, when the constraint reaches the situation where it goes throughpoint B (which is then the optimal point), no further changes are possible. Ifthe constraint were moved further, the optimal point would no longer movesince it cannot go below the x:axis where the y values are negative. Thereforethe optimal point will stay.at B. Therefore no further increases in the resourcelead to improvements in the objective function and consequently the dual valuemust be zero.Point B has y = 0 since it is on the x-axis. Since it is also on the manpower

constraint:

3x + 4y = 903x + 0 = 90

x = 30

5x + 2y = 5 x 30 + 0= 150

Therefore the machine-hours can be increased from 80 to 150 before the dualvalue drops to zero. The right-hand side range for increasing machine-hours is70 (150-80). If extra machine-hours can be made available at a cost less than

the dual value of 0.5, up to 70 of them should be accepted. More than 70 andno further improvements in profit will result.In a similar way the right-hand side range for decreasing machine-hours as

well as for manpower can be calculated. Computer packages produce right-handside ranges automatically. In this case the output would be as follows:

Constraint Initial value Down to Up toMachine-hours 80 45 150Manpower 90 48 160

Profit can be increased by 0.5 (the dual value) for each extra machine-hourup to a total of 150 (including the original 80) and decreased by 0.5 for eachmachine-hour taken away until machine-hours are reduced to 45. Similarly thedual value for manpower applies to each added or subtracted hour up to a totalof 160 or down to 48.Dual values can also be found for the non-negativity constraints, x :::::0, y :::::0,

but, just to be confusing, they are known as the reduced costs of the variables.If the optimal value of a variable is not zero, then its non-negativity constraintis slack and the reduced cost (= dual value) is zero. If, on the other hand, theoptimal value of a variable is zero then the constraint is tight and will have anon-zero reduced cost, which is the amount by which the objective will changeif the variable is forced to be non-zero. For example, suppose the optimal valuefor the productivity level of an electrical component is zero, but the decisionmaker believes that some should be produced in order that a full product rangecan be offered to customers. The reduced cost of the component is the per unitcost of doing so. If the reduced cost were £2 and 1000components were to bemade, then the loss compared to the optimal objective function would be £2000.

Coefficient Ranges

A coefficient range shows by how much an objective function coefficient canchange before the optimal values of the decision variables change. The graphicalsolution to the electronics example (Figure 4.3) helps to show what this means.The optimal point A was found by moving the objective function line outwards(north-easterly) and determining the last point of contact with the feasible region.Had the slope of the objective function been different the optimal point couldhave been B or C. Changing the coefficient values is the same as varying theslope of the line. If they change by small amounts the slope of the line willnot be sufficiently different for the optimal point to move away from A. If theychange by large amounts the slope may be sufficiently different to change theoptimal point. The coefficient ranges are the amounts by which each coefficientof the objective can be varied before the optimal point does move away from Ato B or C. Although small changes in a coefficient may not move the optimalpoint, the optimal objective function value will of course be different.For the electronics example the computer output would be as follows:

Coefficient Initial value Down to Up toProduct X 4 2.25 7.50Product Y 3 1.60 5.33

Objectivefunction

The profit per unit for product X can vary between 2.25 and 7.5 and theoptimal production will still be 10 of X and 15 of y. Similarly the profit per unitof Y can be changed from 3 down to 1.6 and up to 5.33without disturbing theoptimal point. These ranges apply to the variation of one coefficient at a time,the other being kept at its original value.

The following problem was form.ulated and solved in the previous module:

A manufacturer of soft drinks is preparing a production run on two drinks, A and B.There are sufficient ingredients available to make 4000 bottles of A and 8000 of B,but there are only 9000 bottles into which either may be put. It takes 30 minutes toprepare the ingredients to fill 200 bottles of A; it takes 10 minutes to prepare theingredients to fill 200 bottles of B. There are 12 hours available for this operation.The profit is 8p/bottle for A and 6p/bottle for B. The manufacturer wishes to knowhow many bottles of each soft drink he should prepare.

The fully formulated LP problem is:

Max.Subject to

0.08x +x::;y::;

x+y::;3x + Y ::;

x, Y ?:

O.06y400080009000

14400o

A typical computer package used to solve this problem would give output as follows,which includes information about dual values, right-hand side ranges and coefficientranges:

Reducedcost

x 2700 0Y 6300 0

Objective value = 594

Dual RHS rangesConstraint Slack value Original Down to Up to

1 1300 0 4000 2700 Infinity2 1700 0 8000 6300 Infinity3 0 0.05 9000 6400 101334 0 0.01 14400 11000 17000

Variablex

Objective coefficient-rangesOriginal Down to Up to

0.08 0.06 0.180.06 0.027 0.08

(a) The optimal production levels are to make 2700 bottles of drink A and 6300of B.

(b) At these production levels the profit will be £594.

(c) The column headed 'slack' indicates the extent to which each constraint is tightor slack. For example (constraint 1), there are sufficient ingredients for softdrink A left over to make a further 1300 bottles.

(d) The column headed 'Dual value' gives the dual value for each resource orconstraint. As expected a slack constraint has a zero dual value. This is thecase for constraints 1 and 2 referring to the availability of bottles. The thirdconstraint is tight and has a dual value of 0.05. If an extra bottle were available,5p could be added to total profit.

The fourth constraint is also tight, with a dual value of 0.01. The meaning ofthis dual value is harder to interpret because it is not immediately clear what itis that is worth an extra 1p. An extra hour, an extra minute? It is necessary togo back to the constraint itself: 3x + y :::; 14400 . The dual value refers to theincrease in the objective function if the right-hand side of the constraint wereincreased by 1 from 14400 to 14401. Originally this constraint was that the timeavailable on the machine should be less than 12 hours. When the constraintwas formulated, hours were turned into minutes so that the right-hand sidewas 12 x 60. To get rid of some fractions the constraint was multiplied throughby 20 so that the right-hand side became 12 x 60 x 20 = 14400. The unitsare therefore one-twentieths of a minute, i.e. 3 seconds. The dual value meansthat if extra time were available on the machine each extra 3 seconds wouldbe worth 1p. If the extra time could be gained by use of overtime then each

3 seconds would be worth 1p, each minute would be worth 20p, each hourwould be worth 1200p, or £12. It is worth using overtime if the cost is lessthan £12/hour, over and above the cost of normal-time working included in theobjective function.

(e) The columns headed 'RHS ranges' show the boundaries within which the dualvalues apply. Constraints 1 and 2 (ingredients) are slack and therefore they havezero dual values. This would continue to be the case no matter how much extraof the ingredients were available. Hence the right-hand side can be increasedup to 'infinity' without changing the dual values. Likewise the right-hand sidescan be reduced by the amount of the slack and the dual values will remainzero.The dual value is 5p per bottle from 9000 bottles up to 10133 and from 9000down to 6400. For machine time the dual value is 1p per 3 seconds (= £12 perhour). This applies for an additional 2600 x 3 seconds (= 130 minutes) and fora reduction of 3400 x 3 seconds (= 170 minutes).

(f) The section headed 'Objective coefficient ranges' shows that provided the profitfor soft drink A is in the range 6p to 18p per bottle (but assuming the profitfor B is unchanged at 6p), the optimal decision will still be to make 2700 of Aand 6300 of B. Similarly the optimal decision is unchanged while the unit profitof B is in the range 2.7p to 8p (assuming the profit for A stays at 8p).

(g) The reduced cost of both variables is 0, since their optimal values are non-zero.

The LP problems illustrated in these modules would be solved quickly by acomputer package. In some practical applications the size of the LP is so greatthat the time and therefore expense involved in obtaining a solution is animportant factor in deciding how to tackle the problem. For example, the largelinear programmes used by oiLcompanies can have hundreds or thousands ofconstraints and take several-hours to solve even on large mainframe computers.How often to re-run the programme when new data become available is abig issue and it then becomes important to look for short-cut solution methods.Some LP problems have special characteristics which enable them to be solved bydifferent algorithms which take significantly less computer time. Transportationproblems are like this.For example, a cement company manufactures at three different locations:

Alderton, Blackthorne and Chadley. The product is then taken to any of eightdifferent warehouses where it is stored prior to being sold to builders merchants.The cost of transporting cement from factory to warehouse differs of courseaccording to the locations involved. The company's problem is to decide howmuch of each factory's output to send to each of the warehouses so that thecost of distribution is minimised. The problem is represented diagrammaticallyin Figure 4.4.Each factory has a planned monthly output and each warehouse has an esti-

mated demand for cement. For example, Alderton will produce 30000 tormesnext month and warehouse 1 is likely to require 6000 tonnes. These figures

Warehouse1

Factory

Alderton (A)

together with the cost per tonne of moving cement between the different loca-tions are shown in Table 4.1. For example, the cost of transporting one tonne ofcement from factory A to warehouse 4 is £12. The total outputs are exactly equalto the total demands but this does not have to be the case. In a more complicatedproblem provision could be made for storage at factories and warehouses.

Table 4.1 Costs, outputs and demandsOutput Warehouse

(0005 tonnes) 1 2 3 4 5 6 7 8Demand (0005 tonnes) 6 11 8 14 5 1 8 14

Cost (£ per tonne):

Factory A 30 11 43 9 12 25 32 18 56B 18 24 22 76 28 5 43 48 16

C 28 54 14 40 33 38 29 33 8

(a) Decision variables. They must be the tonnages to be transported along eachroute. There are 24 routes (3 factories x 8 warehouses) and therefore 24decision variables:

xAl, xA2, ... , xA8 are the tonnes taken from factory A to warehouses 1,2, ... ,8xBl, xB2, , xB8 are the tonnes taken from factory BxCI, xC2, , xC8 are the tonnes taken from factory C

(b) Objective function. The objective is to minimise the total cost of distributionfor all outputs and demands over all possible routes. The cost per tonne foreach route is taken from the body of Table 4.1:

Minirnise llxAl + 43xA2 + + 56xA8 (factory A)+ 24xBl + + 16x (factory B)+ 54xCl + + 8xC8 (factory C)

(c) Constraints. The first type of constraint says that all the output must gosomewhere, in the absence of a provision for storage. For example, one con-straint is that the 30000 tonnes output of factory A has to go to warehouses.For example:

The output could go to just one warehouse (in which case one variablewould be equal to 30 and the rest 0) or it could be shared in some waybetween some or all of them. There are similar constraints for the other twowarehouses:

xBl + xB2 + + xB8 = 18 (factory B)xCI + xC2 + + xC8 = 28 (factory C)

The second type of constraint is that all demands must be met. For example,a total of 6000 tonnes must be delivered to warehouse 1 from the factories:

The whole 6000 tonnes could come from just one factory or come fromdifferent sources. There are similar constraints relating to the other ware-houses:

xA2 + xB2 + xC2 = 11 (warehouse 2)xA8 + xB8 + xC8 = 14 (warehouse 8)

Apart from non-negativities the 11 above are the full list of constraints. Theproblem could be solved by LP to show how much cement should be sent fromeach factory to each warehouse. In fact the optimal decision is:

xA3= 8 }xA4 = 14xA7 = 8

}xB1 = 6xB5 = 5xB8= 7

xC2 = 11 }xC6 = 10 output from C goes to 2, 6, 8xC8 = 7

This microcomputer solution took just a few seconds but in a larger problemsolution time might be a much more significant consideration. Then, the factthat transportation problems have a special structure which enables them to besolved by a short-cut method and thereby reduce computer time brings abouta worthwhile saving. The special sJructure is that the coefficients of all thevariables in all the constraints have the coefficient '0' or '1'. When a problemhas this characteristic a transportation method algorithm can be used.Usually it is easier to recognise the problem by its nature (transporting from

'factories' to 'warehouses') than from the mathematical formulation. However,not all transportation problems are concerned with taking products from theirplace of manufacture to the point of sale. For example, a regional police authorityhas at least one patrol car attached to each of its eighty police stations. Withinthe area there are three vehicle centres where the cars are serviced and repaired.To which centre should the cars of each station be sent? This is a transportationproblem. The service centres are the 'factories' and their vehicle capacities the'outputs'. The police stations are the 'warehouses' and the numbers of carsattached to them their 'demands'. The cost of servicing a car at one centre ratherthan another is related mainly to the distance between centre and station.Besides transportation problems other situations exhibit special characteristics

which permit a short-cut solution method. Assignment problems are an exam-ple. In a metal-pressing shop there may be five different pieces of work to doand five different presses. The cost of doing each piece of work on each press isshown in Table 4.2.

Table 4.2 Press shop assignment problemPiece of work (cost £)

Press 1 2 3 4 5

A 2.1 3.5 1.8 2.0 4.28 3.0 4.2 3.0 2.7 3.9C 2.5 3.1 2.3 2.3 3.3D 3.5 3.8 3.4 2.9 4.8E 3.4 3.6 2.9 3.1 4.1

The problem is that of assigning each piece of work to a press. When for-mulated as a linear programme this problem has characteristics which allow anassignment algorithm to be used to solve it and thereby reduce computer time

as compared to LP-based methods. An assignment problem is like a transporta-tion problem with the additional feature that the demand at each 'warehouse' isexactly 1 as is the supply at each 'factory'.

An integer is a whole number. For example, I, 2, 3, 4, 5, ... are integers, as are-I, -2, -3/ ... Decimals and fractions, on the other hand, are not.LP assumes that the decision variables are continuous, that is they can take on

any values subject to the constraints, including decimals and fractions. ThereforeLP cannot be applied to problems in which any of the variables must be restrictedto integer values only. An example will show when this might occur.Suppose a shipping company is planning the tonnage it needs in the future.

LP could be used to indicate the tonnage required on each route in order tomaximise future profit subject to a number of constraints on, say, manpower,finance, demand levels, port capacities, etc. It would not matter if the answerswere whole numbers or not. The tonnages are approximate indications of whatis needed and an answer of 63.7 thousand tonnes would be acceptable. On theother hand the problem could have been formulated to decide the number ofships required, not the tonnage. Now it is necessary for the variables to beintegers since it would be difficult to interpret an answer such as 2.5 ships.Integer programming would have to be used to solve this problem.Integer programming uses a different solution method from LP.The algorithm

is more complicated and takes longer to solve by computer. Because of thegreater difficulty and expense in applying integer programming, 'tricks' areused so that LP can be applied. For example, 2.5 ships could be interpreted asowning two ships while chartering one ship for six months of the year.Integer programming also has financial applications. For example, a company

has a series of opportunities for capital investment which have different returnsand cash flow profiles. Only a limited amount of money is, of course, availableand also the company has certain cash-flow requirements through time. Inwhich of the opportunities' should the company invest? The problem can behandled by LP. The decision variables are the amounts of money to invest ineach opportunity; the objective is to maximise return; the constraints are themoney available and the cash-flow requirements through time. Some of theopportunities can have variable amounts invested in them, such as, for example,the purchase of stocks and shares. Other opportunities, however, require all-or-nothing investments, such as the purchase of a motor vehicle or the constructionof a new plant. An answer such as to invest £20000 in new vehicles when thevehicles cost £15000 each would not make sense. The investment must be for a,£15000, £30000, etc. This situation also requires integer programming.

A further extension to basic LP is to non-linear programming. This term refersto formulations where the linearity requirement is relaxed. For example, theobjective function may involve squared terms, allowing for a curved relationship

between profitability and output. Only a few types of non-linear problem can besolved, and those that can be solved generally require a large computer capacity.Quadratic programming is for problems whose objective functions are

quadratic, i.e. squared terms are involved. For example, suppose the quan-tity of a product purchased depends, realistically, on its price. The demandfunction, probably estimated by applying regression analysis to past data ormarket research information, might be:

Quantity purchased = 10000 - 3 x Pricei.e. Q = 10000 - 3P

Revenue = Price x Quantity=P X Q= P(10000 - 3P)= 10 OOOP - 3p2

This is a quadratic equation since if involves P and p2. If such an expressionappeared in an objective function the problem would no longer be one ofLP. Solution methods that assume the objective function is linear (such as thegraphical method used in the previous module) cannot be applied. It is notnecessary, in a managerial context, to investigate the nature of the algorithmrequired since a computer package would always be used. However, for amanager who is likely to need to solve such problems it is important to checkthat the software bought has the capability to handle quadratic programming.Quadratic objective functions might occur in situations where unit costs are

variable. For instance, economies of scale might result in unit costs whichdecrease according to a quadratic equation as the number produced increases.They might also occur in pricing where there are quantity discounts.Determining whether an economy of scale curve or a discount curve isquadratic and, if so, its exact equation, requires the use of estimation meth-ods such as curvilinear regression. This is similar to the way in which linearregression might be used to calculate the demand equation used earlier.Quadratic programming deals with the special case where the objective func-

tion is quadratic but the constraints are still linear. Different programmingmethods have to be used when a problem is non-linear in some other way.Table 4.3 summarises other programming methods and the circumstances inwhich they are used.

Other non-linear programming methodsCharacteristics

ConstraintsObjective function

Linear and squared terms

Both non-linear. Terms are productsof decision variables, e.g. Max.3xy + 4X2y3

Linear

Programming

method

Quadratic

Geometric

Non-linear but can be approximatedby linear expressions

The programming methods used so far assume that all the data are known withcertainty: unit costs, sales demands, resource availabilities, etc., are exact. Inother words the coefficients and constants in the problem formulation are fixed.Such a formulation, or model, is said to be deterministic. In practice, financial,marketing and other information is likely to have come from estimation andforecasting procedures and will therefore be to some degree uncertain. Moreprobably the data will be exceedingly imprecise. Three ways of handling suchuncertainties will be described.The first is by use of sensitivity analysis. This means changing the input dataand re-solving the problem to test the effect of changed data assumptions onthe optimal decision. For example, the right-hand side of a 'minimum sales'constraint might be a demand of 5000. Sensitivity analysis would re-solve theproblem changing the 5000 each time. Carried out exhaustively this will showwhich data have a big influence on the results and which a negligible influence.Some computer packages are able to carry out parametric programming,

which is a systematic form of sensitivity analysis. A coefficient is varied con-tinuously over a range and the effect on the optimal solution displayed. Forexample, in a linear programme to decide the capital expenditure on each ofthree projects, one constraint might impose a limit of £2 million on total expen-diture. It would be of the form:

Investment in project 1+ Investment in project 2

+ Investment in project 3 ::::2000000

Invest in Project 1: £l.5mProject 2: £0Project 3: £O.5m

Parametric programming might vary the £2 million through the range £1millionto £3 million and produce the results in Table 4.4.

Table 4.4Total

investment

limit (L)

1 to 1.81.8 to 2.62.6 to 3

Parametric programming

Investment in

Project

1

L-0.9L-O.52.1

Project

2

0.9oo

Project

3

o0.5

L-2.1

However, parametric programming and other forms of sensitivity analysis donot use any probability information that may be available. For example, marketresearch might have shown demand in the form of a normal distribution, butsensitivity analysis would not use it. It is merely a question of trying differentdemand levels and seeing the effect.The second approach to uncertainty is one which explicitly deals with proba-

bility information. It is called stochastic programming. This is the name for thegeneral class of programming problem which deals with uncertainty by trans-forming the formulation into one that can be handled by ordinary programmingmethods. In the above example, the investment in the projects, if they were thecosts of building new plants, might not be known with certainty. Instead thesedecision variables might be in the form of normal distributions. One way to han-dle this would be to treat the variables as the means of the normal distributionsand then find the best mix of investments on average. Converting an objectivefunction to be an average value is a simple type of stochastic programming.The third approach to uncertainty is chance-constrained programming, whichdeals with the situation where some constraints cannot or do not have to be metall the time but only a certain percentage of the time. For example, in a problemto find the optimal size of a water reservoir, constraints will deal with the levelsof rainfall. One might express the fact that the minimum rainfall in the areais likely to be 40 inches per annum. However, in some exceptional years therainfall might be even lower. If there were a one in twenty chance that it wouldbe lower the constraint would be written in the form:

Rainfall :2 40 (with 95% probability)

Chance-constrained programming can deal with constraints in this form toprovide an answer that has a high probability of being the best.

The extension of LP to methods such as integer and non-linear programminggreatly increases the range of applications. Further extensions are made possibleby the use of 'tricks', meaning methods to get round some of the restrictions onLP.For example, the decision variables are supposed to be greater than or equalto zero: negative numbers are not allowed. However by writing a variable x intwo parts:

x can take on negative values although its two constituent parts, xl and x2,are always greater than or equal to zero. Whenever x should appear in the LPformulation it is substituted by xl - x2 and LP proceeds as normal but of coursethe output has to be interpreted carefully.Other variants on LP allow the decision maker to tackle slightly different types

of problem. For example, goal programming is concerned with setting decisionvariables at levels which bring the decision maker as close as possible to a setof objectives. Goal programming is therefore a type of minimisation problem,minimising the distance between what is feasible and what is being aimed at.

It is an example of the general class of multicriteria methods which handle theproblem of there being more than one objective.Other types of programming method have been mentioned to illustrate the

ways in which LP can be extended and applied to a wider range of circum-stances. Research and development work is continuing and the area is now verylarge and highly specialised. Beyond LP, however, only small problems of thistype (in terms of numbers of decision variables and constraints) are solved inpractice. The reason is that these other methods require much more computerpower than LP. Even with present-day large computers the cost of obtainingsolutions to large problems can be prohibitive. The decision maker has to bevery sure the expenditure is going to be worthwhile.Perhaps a more fundamental problem with the more sophisticated types of

programming is that they are harder for the layman to understand. This maynot be a problem with technical applications but for general management appli-cations it may be difficult for the output to be properly used as a decision aidif users feel intimidated by it. Deciding which is going to be the more effectiverequires a trade-off between an approximate but comprehensible answer and acomplex but more accurate solution.

4.1 Is the following statement true or false?Every decision variable in a linear programme problem has a dual value associ-ated with it.

4.2 Is the following statement true or false?The dual value of a resource is always greater than the price paid for theresource.

Questions 4.3 to 4.6 refer to the following situation:A small engineering company produces two machine parts, L23 and L31. L23 needssix hours of lathe time and three hours on the polishing machine. L31 needs twohours on the lathe and five hours on the polisher. The company has three lathesand three polishers which are worked on a 40-hour week. L23 sells for £140 eachand L31for £220. Variable costs are £100 per unit for both products. Formulated asa profit-maximising linear programme where x and yare the quantities of L23andL31to be produced, the problem becomes:

Max.Subject to

40x + 120y6x + 2y :::;1203x + 5y :::; 120

x:::: 0, y:::: 0

The computer printout of the solution shows:Optimal objective function is 2880Optimal decision variables: x = 0, y = 24

ConstraintLathePolisher

Slack72o

Coefficient rangesL23: 0 to 72L31: 67 upwards

4.3 If an extra hour of polisher time were available what would be the new optimalvalue of the objective function?

A 2880.

B 2890.

C 2904.

D 3000.

4.4 What is the dual value of lathe time?

A O.B 24.

C 40.

D 120.

4.5 A subcontractor offers the use of his polishers, making a total of 200 additionalhours available. By how much could the profit be increased?

A 1728.

B 1800.

C 4320.

D 4800.

4.6 A re-calculation of the profitability of L23 and L31 by the Finance Departmentshows that the true profit per unit of L31 is £100 instead of £120. The objectivefunction should therefore really be 40x + 100y. What is the new optimal valueof the objective function?

A It is unchanged.

B 2400.

C 2640.

D It is impossible to say without completely re-solving the LP problem.

4.7 Is the following statement true or false?

A transportation problem is a type of LP problem with special characteristicsthat make it easier to solve.

4.8 Which of the following statements about an integer programming problem iscorrect?

A The objective function can only take whole number values.

B The right-hand sides of the constraints can only be whole numbers.

C All the decision variables can take only whole number values.

D At least one of the decision variables can take only whole number values.

4.9 The following statements which apply to a programming problem make it anon-linear problem:

I The objective function includes squared decision variables.

II The right-hand side of the constraints are logarithms.

III Some variables appear in the constraints in logarithmic form.

Which of the following is correct?

A I only.

B I and III only.

C II and III only.

D I, II and III.

4.10 Programming techniques other than LP tend not to be greatly used because:

I They are time-consuming and expensive to solve.

II Computers are not powerful enough to solve them.

Which of the following is correct?

A I only.

B II only.

C Both I and II.

D Neither I nor II.

A furniture company's two major product lines are chairs and tables. They arebought in kit form from subcontractors and then assembled and finished in thecompany's factory. Production of a chair requires four hours of assembly timeand two hours of finishing time. The corresponding times for a table are twohOloISand five hours. There are 80 hours of assembly time available per weekand 120hours of finishing time. The profit contribution (revenue minus variablecosts) for a chair is £60 and for a table £80.

1 What production levels per week will maximise total profit contribution?Extra finishing time can be made available by using overtime, which willcost £8 per hour in addition to normal payment. Should overtime be used?

A farmer has to decide the correct amount of fertiliser to apply to his rootvegetable fields. He has been advised by a soil analyst that each acre will requirethe following minimum amounts: 60kg nitrogen compounds, 24kg phosphoruscompounds and 40kg potassium compounds. There are two major types offertiliser available. The first is JDJ which comes in 40kg bags at £6 per bag andwhich comprises 20%nitrogen compounds, 5%phosphorus and 20%potassium.The second, PRP, comes in 60kg bags at £5 each and comprises a 10%-10%-5%mixture. Formulate the problem as a linear programme which minimises cost.

1 How many bags of each product should he use per acre?Being a cautious farmer he wishes to have a second opinion from anothersoil analyst. The second analyst suggests 9kg less phosphorus compoundsare required per acre, so how much will the farmer save per acre? What isthe dual value for phosphorus?

This case study is a continuation of Case Study 3.3 in the previous module.A timber company wants to make the best use of its wood resources in one

of its forest regions which contains fir and spruce trees. During the next month45000 board-feet of spruce are available and 80000 board-feet of fir.The company has mills to produce lumber or plywood. Producing 1000board-

feet of lumber requires 1200 board-feet of spruce and 2500 board-feet of fir.Producing 1000 square feet of plywood requires 2200 board-feet of spruce and3500board-feet of fir.Firm orders mean that at least 6000-board-feetof lumber and 12000 square feet

of plywood must be produced during the next month. The profit contributionsare £1250 per 1000 board-feet of lumber and £2250 per 1000 square feet ofplywood.The production decision is taken on the basis of the following LP model:

Max. 1250L+ 2250PSubject to 1.2L+ 2.2P ::::;45 (spruce)

2.5L + 3.5P ::::;80 (fir)L ~ 6 (orders)P ~ 12 (orders)

L = amount of lumber produced (in ooos board-feet)P = amount of plywood produced (in ooos square feet).

The computer output for this problem is of the form shown in Table 4.5.

Table 4.5

Variable

Lp

Variable Value

L 14.23P 12.69Objective value = 46340Slack Dual price

o 961.5o 38.58.23 00.69 0

Objective coefficient ranges

1230 to 16101750 to 2290

RHS range

44.6 to 48.175.1 to 80.7

Constraint

1

2

3

4

Assuming the constraints are in the same order as in the formulation (e.g.constraint 1 refers to spruce):

1 How much lumber and how much plywood should be produced?

2 How much extra profit contribution could result if the amount of sprucethat arrives at the mills is found to be 45500 board-feet?

3 How much more spruce and fir could be accepted at the mill (assumingthere is sufficient processing capacity) before the extra profit contributiondrops to zero?

4 What would be the optimal production mix if the profit contribution ofplywood were £2290?